Find an example of a bounded domain (open, connected set) $U$ with smooth boundary $\partial U$ that is not convex but is Levi-pseudoconvex and hence show that your claim is true.
My attempt: I thought I should pick a domain of holomorphy in $\mathbb{C}$ that is not convex (because I know that should work), so I settled on attempting to show the set $\{ z\in \mathbb{C} \ | \ 1/2 < |z| < 1 \}$ is pseudoconvex. However, I'm having a bit of difficulty in doing that so I must be going wrong somewhere.
For reference the definition of defining function I'm using is: Given a domain $U$ with smooth boundary a smooth function $r$ is said to be a defining function at a point $p \in \partial U$ if $r$ is defined in a neighbourhood $V$ of $p$ with non-vanishing derivative such that $U \cap V = \{ x \in V \ | \ r(x) = 0 \}$ and $r < 0$ in $U$.
Furthermore, a vector $\sum_{j=1}^{n} a_j \frac{\partial}{\partial z_j}\big\lvert_{p}$ is in $T_p^{(1,0)} \partial U$ if $\sum_{j=1}^{n} a_j \frac{\partial r}{\partial z_j}\big\lvert_{p} = 0$, where $a_j$ are complex numbers.
The definition of Levi-pseudoconvex I'm using is the following.
Suppose $U \subset \mathbb{C}^n$ is an open set with smooth boundary, and $r$ is a defining function for $\partial U$ at $p \in \partial U$. If \begin{equation*} \sum_{j=1,l=1}^{n} \overline{a_j}a_l \frac{\partial^2 r}{\partial \overline{z_j}\partial z_l}\big\lvert_{p} \geq 0 \ \ \text{ for all } \ \ \sum_{j=1}^{n} a_j \frac{\partial}{\partial z_j}\big\lvert_{p} \in T_p^{(1,0)} \partial U \end{equation*} then $U$ is said to be Levi-pseudoconvex at $p$.
For the subset of the boundary of $U$ consisting of $\{ z \in \mathbb{C} \ | \ |z| = 1\} =: A$, use the defining function $r_A(z,\overline{z}) = z\overline{z}-1$. Note that this is a defining function for every $p \in A$, and is negative on $U$. For the other subset of the boundary $\{z \in \mathbb{C} \ | \ |z| = 1/2 \} =: B$ use the defining function $r_B(z,\overline{z}) = \frac{1}{4} - z\overline{z}$. Note that $r_B$ satisfies the above requirements and hence is a defining function for all $p \in B$.
However, now I get stuck. I can't figure out what vectors are in $T_p^{(1,0)}\partial U$ for either of my defining functions. My thinking is that it should just be the zero vector because you have very little freedom here. You want $a \times \overline{z}|_p = 0$ for both, but $p \neq 0$ so $a$ must be zero. Is this correct?
In one dimension (in $\mathbb{C}$) $T_p^{(1,0)} \partial U = \{ 0 \}$, so you are right there. For something to contain a holomorphic tangent vector you need at least two (real) dimensions, and a boundary of a domain in $\mathbb{C}$ is one (real) dimensional. So any $U \subset \mathbb{C}$ is Levi-pseudoconvex. So if you are willing to do this just in $\mathbb{C}$ then the annulus is an example.
If you need an example in higher dimensions, I would suggest a different tack. Start with something that you know is Levi-psuedoconvex, say the ball, and then think of a mapping that biholomorphically takes the ball onto a nonconvex set.